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[SOLVED] Solution of the Damped Wave Equation under Certain Boundary Conditions

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
u_{tt} + 3u_t = u_{xx}\Rightarrow \varphi\psi'' + 3\varphi\psi' = \varphi''\psi.
$$
$$
u(0,t) = u(\pi,t) = 0
$$
$$
u(x,0) = 0\quad\text{and}\quad u_t(x,0) = 10
$$
\[\varphi(x) = A\cos kx + B\sin kx\\\]
\begin{alignat*}{3}
\psi(t) & = & C\exp\left(-\frac{3t}{2}\right)\exp\left[t\frac{\sqrt{9 - 4n^2}}{2}\right] + D\exp\left(-\frac{3t}{2}\right)\exp\left[-t\frac{\sqrt{9 - 4n^2}}{2}\right]
\end{alignat*}
The general sol would be
\begin{eqnarray}
u(x,t)&=&\exp\left[-\frac{3t}{2}\right]\sin x\left[A_1\cosh\frac{t\sqrt{5}}{2} + B_1\sinh\frac{t\sqrt{5}}{2}\right]\\

&+&\exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}\sin nx\left[C_n\cos t\frac{\sqrt{4n^2 - 9}}{2} + D_n\sin t\frac{\sqrt{4n^2 - 9}}{2}\right]

\end{eqnarray}
Correct?
 
Last edited by a moderator:

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
u_{tt} + 3u_t = u_{xx}\Rightarrow \varphi\psi'' + 3\varphi\psi' = \varphi''\psi.
$$
$$
u(0,t) = u(\pi,t) = 0
$$
$$
u(x,0) = 0\quad\text{and}\quad u_t(x,0) = 10
$$
\[\varphi(x) = A\cos kx + B\sin kx\\\]
\begin{alignat*}{3}
\psi(t) & = & C\exp\left(-\frac{3t}{2}\right)\exp\left[t\frac{\sqrt{9 - 4n^2}}{2}\right] + D\exp\left(-\frac{3t}{2}\right)\exp\left[-t\frac{\sqrt{9 - 4n^2}}{2}\right]
\end{alignat*}
The general sol would be
\begin{eqnarray}
u(x,t)&=&\exp\left[-\frac{3t}{2}\right]\sin x\left[A_1\cosh\frac{t\sqrt{5}}{2} + B_1\sinh\frac{t\sqrt{5}}{2}\right]\\

&+&\exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}\sin nx\left[C_n\cos t\frac{\sqrt{4n^2 - 9}}{2} + D_n\sin t\frac{\sqrt{4n^2 - 9}}{2}\right]

\end{eqnarray}
Correct?
Assuming that the gen soln is correct. Here is what I did to solve for the coefficients. Is this correct?
I haven't been able to solve for $B_1$ though. Hopefully, someone will have some insight.
Using the first boundary condition, we have
\begin{alignat*}{5}
u(x,0) & = & A_1\sin x + \sum_{n = 2}^{\infty}C_n\sin nx & = & 0\\
& \Rightarrow & \sum_{n = 2}^{\infty}C_n\sin nx & = & -A_1\sin x\\
& \Rightarrow & C_n & = & -\frac{2A_1}{\pi}\int_0^{\pi}\sin x\sin nxdx\\
& & & = & \frac{2A_1\sin n\pi}{\pi(n^2 - 1)}\\
& & & = & 0
\end{alignat*}
That is, $C_n = 0$.
\begin{alignat*}{3}
u(x,t) & = & \exp\left[-\frac{3t}{2}\right]\sin x\left[A_1\cosh\frac{t\sqrt{5}}{2} + B_1\sinh\frac{t\sqrt{5}}{2}\right] + \exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}D_n\sin nx\sin t\frac{\sqrt{4n^2 - 9}}{2}
\end{alignat*}
Again, using the first boundary condition, we have (Is it okay to use the BC twice?)
\begin{alignat*}{3}
u(x,0) & = & A_1\sin x & = & 0.
\end{alignat*}
Therefore, $A_1 = 0$ too.
\begin{alignat*}{3}
u(x,t) & = & \exp\left[-\frac{3t}{2}\right]B_1\sin x\sinh\frac{t\sqrt{5}}{2} + \exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}D_n\sin nx\sin t\frac{\sqrt{4n^2 - 9}}{2}
\end{alignat*}
Using the second boundary condition, we have
\begin{alignat*}{5}
u_t(x,0) & = & \frac{\sqrt{5}}{2}B_1\sin x + \sum_{n = 2}^{\infty}D_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx & = & 10\\
& \Rightarrow & \sum_{n = 2}^{\infty}D_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx & = & 10 - \frac{\sqrt{5}}{2}B_1\sin x\\
& \Rightarrow & D_n & = & \frac{4}{\pi\sqrt{4n^2 - 9}}\int_0^{\pi}\left(10 - \frac{\sqrt{5}}{2}B_1\sin x\right)\sin nxdx\\
& & & = & -\frac{40(\cos n\pi - 1)}{n\pi\sqrt{4n^2 - 9}}\\
& & & = & -\frac{40((-1)^n - 1)}{n\pi\sqrt{4n^2 - 9}}\\
& & & = & \begin{cases}
0, & \text{if n is even}\\
\frac{80}{n\pi\sqrt{4n^2 - 9}}, & \text{if n is odd}
\end{cases}
\end{alignat*}
\begin{alignat*}{3}
u(x,t) & = & \exp\left[-\frac{3t}{2}\right]B_1\sin x\sinh\frac{t\sqrt{5}}{2}\\
& + & \frac{80\exp\left[-\frac{3t}{2}\right]}{\pi}\sum_{n = 2}^{\infty}\frac{1}{(2n - 1)\sqrt{\left(n^2 -\frac{1}{2}\right)^2 - \frac{9}{16}}}\sin nx\sin t\frac{\sqrt{\left(n^2 -\frac{1}{2}\right)^2 - \frac{9}{16}}}{2}
\end{alignat*}
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi dwsmith, :)

Can you please explain how you got,

\[\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\]

\[\Rightarrow C_n = -\frac{2A_1}{\pi}\int_0^{\pi}\sin x\sin nxdx\]
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Hi dwsmith, :)

Can you please explain how you got,

\[\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\]

\[\Rightarrow C_n = -\frac{2A_1}{\pi}\int_0^{\pi}\sin x\sin nxdx\]
Fourier coefficient
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Fourier coefficient
Note that when you substitute \(t=0\) in,

\begin{eqnarray} u(x,t)&=&\exp\left[-\frac{3t}{2}\right]\sin x\left[A_1\cosh\frac{t\sqrt{5}}{2} + B_1\sinh\frac{t\sqrt{5}}{2}\right]\\ &+&\exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}\sin nx\left[C_n\cos t\frac{\sqrt{4n^2 - 9}}{2} + D_n\sin t\frac{\sqrt{4n^2 - 9}}{2}\right] \end{eqnarray}

you get,

\[\sum_{n = 2}^{\infty}C_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx = -A_1\sin x\]
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Note that when you substitute \(t=0\) in,

\begin{eqnarray} u(x,t)&=&\exp\left[-\frac{3t}{2}\right]\sin x\left[A_1\cosh\frac{t\sqrt{5}}{2} + B_1\sinh\frac{t\sqrt{5}}{2}\right]\\ &+&\exp\left[-\frac{3t}{2}\right]\sum_{n = 2}^{\infty}\sin nx\left[C_n\cos t\frac{\sqrt{4n^2 - 9}}{2} + D_n\sin t\frac{\sqrt{4n^2 - 9}}{2}\right] \end{eqnarray}

you get,

\[\sum_{n = 2}^{\infty}C_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx = -A_1\sin x\]
$0\cdot\frac{\sqrt{4n^2 - 9}}{2} =0$ and the cosine of 0 is 1
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$0\cdot\frac{\sqrt{4n^2 - 9}}{2} =0$ and the cosine of 0 is 1
Ah, I was confused by the lack of parenthesis. I thought that \(\cos t\) was multiplied by \(\frac{\sqrt{4n^2 - 9}}{2}\).

Anyway it is incorrect that,

\[\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\]

\[\Rightarrow C_n = -\frac{2A_1}{\pi}\int_0^{\pi}\sin x\sin nxdx\]

Note that the Fourier series of the sine function is zero and hence does not converge to the function itself.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Ah, I was confused by the lack of parenthesis. I thought that \(\cos t\) was multiplied by \(\frac{\sqrt{4n^2 - 9}}{2}\).

Anyway it is incorrect that,

\[\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\]

\[\Rightarrow C_n = -\frac{2A_1}{\pi}\int_0^{\pi}\sin x\sin nxdx\]

Note that the Fourier series of the sine function is zero and hence does not converge to the function itself.
I have that $C_n = 0$. Can I use the same boundary condition to obtain $A_1$? How can I obtain $B_1$?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Ah, I was confused by the lack of parenthesis. I thought that \(\cos t\) was multiplied by \(\frac{\sqrt{4n^2 - 9}}{2}\).
This is precisely why I always insist that students put parentheses around function arguments. Don't write so that you can be understood. Write so you can't be misunderstood.

$$ \cos \left(t \cdot \frac{ \sqrt{4n^{2}-9}}{2} \right)$$
is better than
$$ \cos t \cdot \frac{ \sqrt{4n^{2}-9}}{2}\;\text{or}\;\cos t \frac{ \sqrt{4n^{2}-9}}{2},$$
because the latter is ambiguous.

It might be a tad more typing up front, but it saves typing later.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I have that $C_n = 0$. Can I use the same boundary condition to obtain $A_1$? How can I obtain $B_1$?
\(C_n =A_{1}= 0\) is a trivial solution of the equation, \(\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\). You don't need to use a boundary condition for the second time to say that. However it may not be the only solution.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
\(C_n =A_{1}= 0\) is a trivial solution of the equation, \(\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\). You don't need to use a boundary condition for the second time to say that. However it may not be the only solution.
How do I find $B_1$?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I think the solution you have obtained for the damped wave equation is incorrect. Refer, >>this<<.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I think the solution you have obtained for the damped wave equation is incorrect. Refer, >>this<<.
There equation doesn't account for overdamped solutions. If they let $c = 5$, the first few n terms would be over damped and they would need a summation cosh and sinh + summation of cos and sine.

In my case, I have overdamping at n = 1. So I don't need a summation. I just have one term.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I think the solution you have obtained for the damped wave equation is incorrect. Refer, >>this<<.
This linked lead me to another pdf from that school that helped with another problem though :)
 

dwsmith

Well-known member
Feb 1, 2012
1,673
\(C_n =A_{1}= 0\) is a trivial solution of the equation, \(\sum_{n = 2}^{\infty}C_n\sin nx = -A_1\sin x\). You don't need to use a boundary condition for the second time to say that. However it may not be the only solution.
If it isn't the only solution, how can we find another solution?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I have a form for $B$ but I not to sure about it
$$
B=\frac{4 \sqrt{5} \text{Csc}[x] \left(\pi -2 i \text{ArcTanh}\left[e^{-i x}\right]+2 i \text{ArcTanh}\left[e^{i x}\right]+4 \text{Sin}[x]\right)}{\pi }
$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I have a form for $B$ but I not to sure about it
$$
B=\frac{4 \sqrt{5} \text{Csc}[x] \left(\pi -2 i \text{ArcTanh}\left[e^{-i x}\right]+2 i \text{ArcTanh}\left[e^{i x}\right]+4 \text{Sin}[x]\right)}{\pi }
$$
When I plot my soln, I don't see any damping. Is there a mistake some where?
Using the second boundary condition, we have
\begin{alignat*}{5}
u_t(x,0) & = & \frac{\sqrt{5}}{2}B_1\sin x + \sum_{n = 2}^{\infty}D_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx & = & 10\\
& \Rightarrow & \sum_{n = 1}^{\infty}d_n\frac{\sqrt{4n^2 - 9}}{2}\sin nx & = & 10\\
& \Rightarrow & d_n & = & \frac{40}{\pi\sqrt{4n^2 - 9}}\int_0^{\pi}\sin nx dx\\
& & & = & -\frac{40(\cos n\pi - 1)}{n\pi\sqrt{4n^2 - 9}}\\
& & & = & \begin{cases}
0, & \text{if n is even}\\
\frac{80}{n\pi\sqrt{4n^2 - 9}}, & \text{if n is odd}
\end{cases}
\end{alignat*}
If we peel off the $n = 1$ term now, we will have the first term which is $\frac{40}{\pi}\sin x$.
So $B_1 = \frac{16\sqrt{5}}{\pi}$
\begin{alignat*}{3}
u(x,t) & = & \exp\left[-\frac{3t}{2}\right]\frac{16\sqrt{5}}{\pi}\sin x\sinh\frac{t\sqrt{5}}{2} + \frac{80\exp\left[-\frac{3t}{2}\right]}{\pi}\sum_{n = 2}^{\infty}\frac{\sin nx\sin\left(t\frac{\sqrt{4(2n - 1)^2 - 9}}{2}\right)}{(2n - 1)\sqrt{4(2n - 1)^2 - 9}}
\end{alignat*}